Math 1B worksheet
Oct 5, 2009
1{9. Determine whether the following series converge or diverge. (In problem 3, determine, for which integer values of k the series converges.)
∞
X
sin2 n
,
n2 + 1
(1)
n+1
,
n2 + 1
(2)
2 + 3 nk
,
3 + 2 n2k
(3)
sin(1/n2 ),
(4)
n=1
∞ p
X
n=1
∞
X
n=1
∞
X
n=1
∞
X
n=1
n + 4n
,
n + 6n
p
∞
X
n+1−
n
n=1
∞
X
(5)
p
n
,
(6)
1 − cos(1/n),
(7)
p
n=1
∞
X
n=1
1
1+
n
2
∞
X
1
.
n!
n=1
1
e−n ,
(8)
(9)
Hints and answers
We write an ∼ bn if limn→∞ abnn exists and lies strictly between 0 and ∞
(in other words, if we may applyPthe limit comparison test).
2
1
n
1
1. We have sin
and ∞
n=1 n2 converges.
n2 +1
n2
Answer: Converges.
p
1
p
2. We have nn+1
2 +1 ∼ n n .
Answer: Converges.
2+3nk
. For k < 0, we have limn→∞ an = 32 , so the series
3. Let an = 3+2
n2k
diverges by Test for Divergence. For k = 0, we have an = 1, so the series also
diverges. For k > 0, we have an ∼ n1k .
Answer: Converges for k > 1, diverges for all other k.
4. We have sin(1/n2 ) ∼ n12 since limn→∞ n12 = 0 and limx→0 sinx x = 1.
Answer: Converges.
5. We write n+4n = 4n (1+ 4nn ) and limn→∞ 4nnP= 0; therefore, n+4n ∼ 4n .
4n
Similarly, n+6n ∼ 6n , so our series is equivalent to ∞
n=1 6n , a geometric series.
Answer: Converges.
p
p
6. Divide and multiply by n + 1 − n to get that our series is equivalent
P∞
to n=1 np1 n .
Answer: Converges.
2
p 7. We have 1 − cos(1/n) = 2 sin (1/(2n)) by double angle formula, so
1 − cos(1/n) ∼ sin(1/(2n)) ∼ 1/n.
Answer: Diverges.
8. We
have limn→∞ (1+1/n)2 = 1, so our series is equivalent to the geometric
P∞
series n=1 e−n .
Answer: Converges.
9. We write n! = 1 2 n and replace all the terms except the rst one
1
1
by 2, to get n! 2n−1 . So, n!
; the latter is a geometric series. (Many
2n−1
1
1
more ways to compare exist; for example, n!
n(n−1) for n 2.)
Answer: Converges.
2